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 May 8th, 2018, 05:48 AM #1 Senior Member   Joined: Nov 2011 Posts: 239 Thanks: 2 Matrix What is the connection between matrices and linear transformation?
May 8th, 2018, 08:14 AM   #2
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 Originally Posted by shaharhada What is the connection between matrices and linear transformation?

 May 9th, 2018, 05:59 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 This really should be in "Linear Algebra" rather than "Abstract Algebra". A "linear transformation", T, from vector space U to vector space V, is a function taking vectors in U to vectors in V with the property that if $\displaystyle u_1$ and $\displaystyle u_2$ are vectors in U and a and b are scalars, the $\displaystyle T(au_1+ bu_2)= aT(u_1)+ bT(u_2)$. It is easy to show that an m by n matrix is a linear transformation from the vector space $\displaystyle R^n$ to the vector space $\displaystyle R^m$. What makes matrices especially useful is that any linear transformation from a vector space with dimension n to a vector space with dimension m can be represented by a matrix. Let U be the vector space of dimension n. Then there exists a basis $e_1$, $e_2$, ..., $e_n$. A vector in U can be written as a linear combination of those vectors: $a_1e_1+ a_2e_2+ ...+ a_ne_n$. We can represent that as the column matrix $\begin{bmatrix}a_1 \\ a_2 \\ \cdot\cdot\cdot \\ a_n\end{bmatrix}$. Let V be the vector space of dimension n. Then there exists a basis $f_1$, $f_2$, ..., $f_n$. A vector in V can be written as a linear combination of those vectors: $b_1f_1+ b_2f_2+ ...+ b_nf_n$. We can represent that as the column matrix $\begin{bmatrix}b_1 \\ b_2 \\ \cdot\cdot\cdot \\ b_n\end{bmatrix}$. Apply the linear transformation, T, to basis vector $e_1$. $T(e_1)$ is a vector in V so can be written as a linear combination of basis vectors for V: $b_1f_1+ b_2f_2+ ...+ b_nf_n$. The first column of the matrix representation of T is $\begin{bmatrix}b_1 \\ b_2 \\ \cdot\cdot\cdot \\ b_n\end{bmatrix}$, the column matrix corresponding to that vector. Applying T to basis vector $e_2$ gives the second column, to $e_3$ the third column, etc. For example, suppose U is the vector space of polynomials of degree two or less, $ax^2+ bx+ c$, V is $R^2$, and T is the linear transformation that maps $ax^2+ bx+ c$ to $\begin{bmatrix}a \\ b+ c\end{bmatrix}$. Take $x^2$, $x$, and 1 as a basis for U and $\begin{bmatrix}1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix}0 \\ 1 \end{bmatrix}$ as basis for V. Applying T to $x^2$ gives $\begin{bmatrix}1 \\ 0 \end{bmatrix}$, applying T to $x$ gives $\begin{bmatrix}0 \\ 1 \end{bmatrix}$, and applying T to $1$ gives $\begin{bmatrix}0 \\ 1 \end{bmatrix}$. Using those as column, the matrix representation of T, in these bases, is $\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1\end{bmatrix}$. That is, to apply T to $ax^2+ bx+ c$ we write it as $\begin{bmatrix}a \\ b \\ c \end{bmatrix}$ and do the matrix multiplication $\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix}a \\ b \\ c \end{bmatrix}= \begin{bmatrix}a \\ b+ c\end{bmatrix}$ Thanks from topsquark Last edited by skipjack; May 9th, 2018 at 08:25 PM.

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